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18 votes
1 answer
682 views

De Rham via topoi

Étale cohomology of schemes $X$ is constructed as follows: one associates to $X$ the so-called étale topos of $X$, and then one just takes the sheaf cohomology of that topos. Is it possible to ...
user1022117's user avatar
15 votes
1 answer
2k views

Artin vanishing for Stein manifolds and restriction maps

In the setting of complex Stein manifolds $X$ of complex dimension $d$, the theorem of Andreotti--Frankel implies the vanishing of the singular cohomology group $H^i(X,\mathbb Z)=0$ for $i>d$. With ...
Peter Scholze's user avatar
13 votes
1 answer
725 views

Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex?

If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex ...
ychemama's user avatar
  • 1,346
12 votes
0 answers
301 views

Is there a differential form which corresponds to an eigenvalue of the homomorphism in cohomology?

Let $M$ be a closed manifold and $f:M\to M$ be a diffeomorphism. Suppose the homomorphism $f^*:H^k(M;\mathbb R)\to H^k(M;\mathbb R)$ has an eigenvalue $\lambda\in\mathbb{R}$. Note that $\lambda$ is ...
Andrey Ryabichev's user avatar
9 votes
2 answers
1k views

Hodge dual of de Rham cohomology and singular cohomology

We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology?...
wonderich's user avatar
  • 10.5k
9 votes
1 answer
392 views

Integrating hypercohomology classes

Let $X$ be a complex variety. By Poincare's lemma, its singular cohomology can be computed as hypercohomology of the holomorphic de Rham complex (viewing $X$ as a complex manifold) $$\text{H}^\cdot(X,\...
Pulcinella's user avatar
  • 5,701
9 votes
0 answers
347 views

Is there a Hodge isomorphism theorem for part-tangential, part-normal, harmonic differential forms?

Let $M$ be an oriented compact Riemannian $n$-manifold with boundary $\partial M$. A differential $p$-form $\omega$ on $M$ is normal if $i^* \omega = 0$ holds, tangential if $i^* \star \omega = 0$ ...
Enok's user avatar
  • 91
7 votes
1 answer
2k views

What is the scope of validity of Kunneth formula for de Rham?

In books like Bott-Tu or all pdf texts I have found on internet, the Kunneth formula for manifolds $M$ and $N$ and their de Rham cohomology $$ H^{\bullet}_{dR}(M \times N) \simeq H^{\bullet}_{dR}(M) \...
ychemama's user avatar
  • 1,346
6 votes
1 answer
2k views

Integration currents vs Poincaré dual

Let $M$ be a complex manifold of dimension $n$ and $S \subset M$ a closed complex submanifold of complex codimension $r$. Let $[S] \in H_{2r}(S)$ be the fundamental class of $S$. We have the ...
Student85's user avatar
  • 151
6 votes
1 answer
374 views

De Rham cohomology of Lie groupoid

Let $G$ be a Lie group acting on a manifold $M$. Consider the transformation groupoid $\mathcal{G}=(G\times M\rightrightarrows M)$. We have the notion of de Rham cohomology of a Lie groupoid by ...
Praphulla Koushik's user avatar
5 votes
1 answer
228 views

Which compact (finite dimensional) Lie groups have $H^1_{DR}(G)\neq 0$

In particular, I am wondering if $H^1_{DR}(G)\neq 0$ implies that the group can written as a semidirect product of $\mathbb{S^1}$ and something else, with the $\mathbb{S^1}$ factor being responsible ...
R Mary's user avatar
  • 979
5 votes
2 answers
2k views

Global Definition of the Dolbeault Complex of a Vector Bundle

For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from ...
Jean Delinez's user avatar
  • 3,399
5 votes
0 answers
304 views

Generalization of de Rham cohomology, or cohomology for non-smooth case

Let $\Omega\subseteq \mathbb{R}^{3}$ be a contractible region and $f:\Omega\to \mathbb{R}^{3}$ be a vector field on $f$. If $f$ is smooth vector field and $\nabla\cdot f=0$, then $f=\nabla\times g$ ...
Seewoo Lee's user avatar
  • 2,215
4 votes
0 answers
173 views

Spencer complex and de Rham Complex

in those lectures notes written by Claude Sabbbah: https://perso.pages.math.cnrs.fr/users/claude.sabbah/livres/sabbah_nankai110705.pdf there is the proposition 1.4.4 where he says that there is a ...
Pierre21's user avatar
  • 385
4 votes
0 answers
109 views

Generalized de Rham cohomology on product bundle giving specified cohomology

Given a compact, smooth manifold $M$ and a real vector bundle $E \to M$ (in general not flat). There already have been numerous questions about how to equip the space $\bigoplus_k \Gamma(\Lambda^k T^* ...
user avatar
4 votes
0 answers
441 views

Finite Field Varieties and the de Rham Complex of Kähler Differentials

In an answer to a previous question I asked, where the Kahler differentials of a variety over a finite field were discussed, it was stated that: You can certainly define de Rham cohomology using ...
Jean Delinez's user avatar
  • 3,399
3 votes
1 answer
329 views

Calculation of Frobenius on de Rham cohomology of elliptic curves with good reduction

I'm reading "An introduction to the theory of $p$-adic representations" by Laurent Berger. In the page 14 it says: Suppose $F$ is an unramified finite extension of $\mathbb Q_p$ and $E$ is ...
Richard's user avatar
  • 775
3 votes
1 answer
167 views

Models for computing cohomology of Lie groupoids

Given a Lie groupoid $\mathcal{G}=[\mathcal{G}_1\rightrightarrows \mathcal{G}_0]$, let $\mathcal{G}_\bullet$ be the associated simplicial manifold. Let $\Omega^\bullet(\mathcal{G}_\bullet)$ be the ...
Praphulla Koushik's user avatar
3 votes
0 answers
217 views

Reference: Relative cohomology of a morphism

Let $f\colon Y \to X$ be a morphism of schemes, the inverse image in $K$-theory always fit into a long exact sequence $$ \cdots \to K_i(f)\to K_i(X) \xrightarrow {f^*} K_i(Y)\to \cdots $$ where the ...
Tintin's user avatar
  • 2,871
2 votes
1 answer
366 views

Easier ways to compute homology/cohomology by adding extra structure

Suppose $X$ is a topological space and I want to talk about its “homology”. There is this notion of singular homology obtained from singular chain complex. This is not very easy to compute. Suppose ...
Praphulla Koushik's user avatar
2 votes
0 answers
235 views

Mayer-Vietoris on Fibered Products

Suppose $M \xrightarrow{p} X$ is a surjective submersion of smooth manifolds. Define the fibered product in this case: $$M \times_X M = \{ (m_1, m_2) \in M \times M | p(m_1) = p(m_2) \}$$ and let $U =...
cheyne's user avatar
  • 1,611
1 vote
0 answers
303 views

Berthelot-Ogus comparison isomorphism

On the link, page, $ 2 $, the Berthlot-Ogus isomorphism theorem is stated as follows, We have a canonical isomorphism, $$ \rho_{\mathrm{cris}} \ : \ H_{\mathrm{cris}}^{i} (X) \otimes_{K_ {0}} K \to H_{...
Angel65's user avatar
  • 595
1 vote
0 answers
217 views

Find torsion classes using flat bundles

My question refers to a discussion from this older thread on Neron-Severi group of a Kähler manifold. In the comments below Ted Shifrin's answer there arose a discussion when the map $H^2(X,\mathbb{Z}...
user267839's user avatar
  • 6,028
1 vote
0 answers
348 views

Vanishing of cohomology groups

Given a smooth hypersurface $X \subset \mathbb{P}^{2p+1}_{\mathbb{C}}$ of degree $d \ge 5$ (with $p\ge 2$), when can we say that $H^{p+1}(\mathcal{O}_X)=H^{p+1}(\Omega^1_X)=...=H^{p+1}(\Omega^{p-2}...
Naga Venkata's user avatar
  • 1,040
-2 votes
1 answer
89 views

Alternating property of H_2(T, Z)

Let us consider the torus $T = S^1_X \times S^1_Y$, where the former $S^1_X$ has the coordinate $X$, whereas the latter $S^1_Y$ has $Y$. We have an alternate property $dX \wedge dY = - dY \wedge dX \...
Pierre's user avatar
  • 563